I think when the cell is set up as written above (with the diagram) it means
Ag + Cl^- ==> AgCl + e E = -0.222
Cu^2+ + 2e --> Cu E = 0.337
So the sum is +0.115 or c.
Calculate for the electrochemical cell below,
Ag(s) | AgCl(s) | Cl–(aq) || Cu2+(aq) | Cu(s)
given the following standard reduction potentials.
Cu2+(aq) + 2 e– -> Cu(s) E = +0.337 V
AgCl(s) + e– -> Ag(s) + Cl–(aq) E = +0.222 V
a. –0.115 V
b. –0.107 V
c. +0.115 V
d. +0.452 V
e. +0.559 V
Answer A
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